asp.net mvc 4 and the web api pdf free download : Merge pdf online software Library dll windows asp.net web page web forms mechanics35-part1308

10|CoupledOscillators
347
Intermsofthevariables
!
and
k
,theequation(10.37)isthen
!
2
=
4
T
m‘
sin
2
(
=
2)=
4
T
‘
2
sin
2
1
2
k‘
(10
:
41)
Thephasevelocityofthewavesatises
v
2
=
!
2
k
2
=
4
T
‘
2
sin
2
1
2
k‘
k
2
Forsmallvaluesof
k‘=
2,lessthan1/2orso,thesinefunctionisnearlylinear,andthisspeedthenagrees
with
v
2
=
T=
asinthecaseofacontinuousstring,Eq.(7.7). Putthisintermsofthewavelength
anditis
k‘
=2
‘=<
1,or
>
2
‘
. Onewavelengthspansmanymassessothatthesystemstartsto
looklikeacontinuousstring.
Putthewaveitselfintothenotationappropriateforwaves.
y
n
=
e
i
(2
n=N
i!
1
t
=
e
i
(
n
!t
)
=
e
i
(
k‘n
!t
)
=
e
i
(
kx
!t
)
Thisisexactlythestandardformusedinchapterseven.
k
is positive ornegative depending onthe
directionofmotion. ThisisalsowhyIchosetheoscillationtobeproportionalto
e
i!t
insteadof+
i
.
Lookbackatthediscussionabouttheseveraldierentdenitionsofwavevelocity,section7.13,
Eq.(7.62).There,recall,youseethedierencesbetweentwoofthepossibledenitionsofwavevelocity:
phasevelocityandgroupvelocity. Heretheyare
v
phase
=
!
k
=
s
T
sin
1
2
k‘
1
2
k‘
;
and
v
group
=
d!
dk
=
s
T
cos
1
2
k‘
(10
:
42)
Whenthe wavelengthis longenough(
k‘
1), both of these velocities s become
p
T=
. At t much
higherfrequencies,forwhichthewavelengthgetsdownnearthespacing
,thesebecome
2
k‘
=
v
group
v
phase
p
T=
Fig.10.12
Whatis
when
=
?Noticethatthegroupvelocitycanbeintheoppositedirectionfromthephase
velocity.
Therearesomecuriouspuzzlesaboutthissystem. Lookagainattheexampleusedonthelast
fewpages,
N
=10,sothat
=2
s=
10for0
s<N
with
s
aninteger.For
s
=5thisis
=
,and
y
n
=cos(
n
!t
)=cos(
n
)cos(
!t
).ThepictureisFigure10.11,andfromEq.(10.41)youcansee
thatthisisthehighestfrequencyofallthemodesofoscillation.But,thiswaveisnotgoinganywhere;
itisstandingstill.Thisissodespitethefactthatthephasevelocityisnotzero.That’sodd,butnotice
thatthegroupvelocityis zeroatthatpoint. Thisisanotherindicationthatthephasevelocityisnot
asusefulaconceptasyoumayhavebeenledtobelieve. (Itwasprobablytheonly wavevelocitythat
yousawinyourrstintroductiontothesubject.) Thegroupvelocityisnegativeforvaluesof
above
themiddle,andthiscorrespondstothefactdescribedacoupleofparagraphsafterFigure10.10:When
youmovedownfromthemaximum
,yougetaleft-movingwave.
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10|CoupledOscillators
348
10.7PerturbationTheory_
The specialcasesforwhichthe matricesaresimpleandthe problemisreadily solvableareprecisely
that|special. Whataboutthe e othercases forwhichadirectattackwill leadtoamorassofmessy
mathematics?
Thereisonesetofcircumstancesforwhichthereishope.Thatoccurswhenthesystemisalmost
easy. Thatis,ifthematricesareveryclosetoonesforwhichthereisaneasysolution. . (Oratleasta
knownsolution.)
TheideaofperturbationtheoryistowriteMandKeachasthesumoftwoterms:onesimple,
onesmall.Thenwritexand
!
2
asasumofterms,aninniteseriestypically,forwhichIcancompute
therstfewterms. Thiscanprovideagoodapproximatesolutiontothefullproblem.
Foranexampleofthis,taketheprototypeproblem,Eq.(10.1). If
m
1
=1
:
01kg
; m
2
=0
:
98kg
;
k
1
=10
:
0N/m
; k
3
=10
:
2N/m
;
and
k
2
=5
:
0N/m
thenthesystemisalmostsymmetricanditssolutionisclosetothatofthesimple,symmetricsystem.
Buthowclose?
Useanexpansionparameter
asaconvenientwaytokeeptrackofthepowersinanexpansion,
andassumeaseriesexpansionfortheunknownsxand
!
2
.
M=M
0
+
M
1
K=K
0
+
K
1
x=x
0
+
x
1
+
2
x
2
+
!
2
=(
!
2
)
0
+
(
!
2
)
1
+
2
(
!
2
)
2
+
(10
:
43)
SubstitutetheseintotheequationsthatIwanttosolve.
!
2
Mx+Kx=0
(
!
2
)
0
+
(
!
2
)
1
+
2
(
!
2
)
2
+

M
0
+
M
1

x
0
+
x
1
+
2
x
2
+
+
K
0
+
K
1

x
0
+
x
1
+
2
x
2
+
=0
Theonlywaythatthiscanbevalidforarbitrary
isfortherespectivecoecientsofthepowersof
toagree.
(
0
)
(
1
)
(
2
)
(
!
2
)
0
M
0
x
0
+K
0
x
0
=0
(
!
2
)
0
M
0
x
1
(
!
2
)
0
M
1
x
0
(
!
2
)
1
M
0
x
0
+K
0
x
1
+K
1
x
0
=0
(
!
2
)
0
M
0
x
2
(
!
2
)
0
M
1
x
1
(
!
2
)
1
M
0
x
1
(
!
2
)
1
M
1
x
0
(
!
2
)
2
M
0
x
0
+K
0
x
2
+K
1
x
1
=0
(10
:
44)
Itappears thatI’ve madethe problem worse, , notbetter. . I’ve e replaced onedicultequation by an
innitenumberofdicultequations. But t no. . Inwhatlooks s tobe ahopelesstangle, , there’ssome
simplicityhidden. Besides,I’mnevergoingtogetaround d tousingthethirdof these equations,only
thersttwo. Andwhythefrequencysquared insteadofthefrequency? That’sjustthewayitworks,
butseeproblem7.16.Writingthefrequencythiswayisunnecessarilycumbersome,soIwillstopdoing
it.Insteadof(
!
2
)
1
Iwillwrite
!
2
1
.
Equation(
0
) is theequationthatI’vealreadysolved. . It’stheonethatissupposedly y simple.
TherstnewequationthatIhavetoexamineis(
1
).
!
2
0
M
0
x
1
+K
0
x
1
=
!
2
0
M
0
+K
0
x
1
=+
!
2
0
M
1
x
0
+
!
2
1
M
0
x
0
K
1
x
0
(10
:
45)
LookatthewaythatIrearrangedthisequationandyouseethatontheright-handsidetheonlything
thatisunknownis
!
2
1
,thetermthatgivestherstordercorrectiontothefrequency.Ontheleftside
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10|CoupledOscillators
349
youhavesomethingthatlookslikethesimple,unperturbedequation.AtthispointIhavetostopand
presentatheorem.ItissuchaneasytheoremthatyoumaywonderwhyI’dbothercallingitsuch,but
itisthekeytosolvingalltheseequations.Itbreaksopentheentirestructureandmakesitmanageable.
Theorem:If(1)HisamatrixthatsatisesH
y
=
~
H
=H,(Hermitian)
and(2)Hhasanulleigenvector|thereisanxsuchthatHx=0
thenforany columnmatrixy,
x
y
Hy=0
(10
:
46)
Thatis,ifxisaneigenvectorwithzeroeigenvaluethenxisorthogonaltotheoutputofHfor
anyinput.
Theproofinvolveswritingitout.
x
y
Hy=
Hx
y
y=0
(10
:
47)
It’sassimpleasthat.IfyouwanttoseeitwrittenintermsofcomponentsIcandothattoo.
x
y
Hy=
X
i;j
x
i
H
ij
y
j
=
X
i;j
H
ij
x
i
y
j
=
X
i;j
~
H
ji
x
i
y
j
=
X
i;j
H
ji
x
i
y
j
=
Hx
y
y
(10
:
48)
Thematrix ontheleftsideofEq.(10.45)andthatactsonx
1
ispreciselysuchamatrix. The
(
0
)equationsaysthatx
0
isanulleigenvector. ThatmeansthatifImultiplytheequation(
1
)byx
y
0
thenImustgetzero.
x
y
0
!
2
0
M
0
x
1
+K
0
x
1
=0=x
y
0
!
2
0
M
1
x
0
+
!
2
1
M
0
x
0
K
1
x
0
There’snomatrixinversiontodohere,justascalarproduct,andtheresultisthatitdeterminesthe
rstordercorrectiontothefrequency,
!
2
1
.
0=
!
2
1
x
y
0
M
0
x
0
+x
y
0
!
2
0
M
1
x
0
K
1
x
0
so
!
2
1
=x
y
0
!
2
0
M
1
x
0
+K
1
x
0

x
y
0
M
0
x
0
(10
:
49)
Computingthelowestordercorrectiontothefrequency(squared)isnowjustsomematrixmultiplication.
NoticethatwhatshowsupinthedenominatoristhesamescalarproductasinEq.(10.22).
Doesallofthismaterialseemfamiliar,somethingyou’vestudiedbefore?Returntosection7.10
andthedevelopmentofperturbationtheoryforstandingwaves.Therethemathematicsweredierential
equations and heretheyare matrices, , butthe underlyingstructureofthe calculationsarethe same.
Herearesomekeyequationsfrombothcalculationssothatyoucancomparethem.
Eq.(7.45)
=
0
+

1
F
=
F
0
+
F
1
+
2
F
2
+
3
F
3
+
!
2
=
!
2
0
+
!
2
1
+
2
!
2
2
+
Eq.(10.43)
M=M
0
+
M
1
;
K=K
0
+
K
1
x=x
0
+
x
1
+
2
x
2
+
!
2
=
!
2
0
+
!
2
1
+
2
!
2
2
+
Ifyou’vestudiedperturbationtheoryfortheinertiatensorinsection8.7thenagain,youarerepeating
thesamebasicideas. Theequations(8.49)lookverymuchlike(10.43). Thisissucientlyimportant
thatitisworthdoingthreetimes. Maybemore.
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10|CoupledOscillators
350
Eq.(7.47)(b)
TF
00
1
+
0
!
2
0
F
1
1
!
2
0
F
0
0
!
2
1
F
0
Eq.(10.45)
!
2
0
M
0
+K
0
x
1
=+
!
2
0
M
1
x
0
+
!
2
1
M
0
x
0
K
1
x
0
Eq.(7.49)
Z
L
0
dxF
0
(
x
)
TF
00
1
+
0
!
2
0
F
1
=0
Eq.(10.46)
x
y
Hy=0
Eq.(7.52)
!
2
1
!
2
0
Z
L
0
dx
1
(
x
)
F
0
(
x
)
2
0
Z
L
0
dxF
0
(
x
)
2
Eq.(10.49)
!
2
1
=x
y
0
!
2
0
M
1
x
0
+K
1
x
0

x
y
0
M
0
x
0
Example
Takethe caseof two massesand threesprings,Eq. (10.1). . Ifyou u restrictyourselftothespecial
symmetriccaseofforwhich
k
1
=
k
3
and
m
1
=
m
2
thenthemodesandthefrequenciesaresimple.
!
2
0
=
k
1
m
;
x
0
=
1
1
and
!
2
0
=
k
1
+2
k
2
m
;
x
0
=
1
1
(10
:
50)
Whathappens ifIstickasmallwadof chewinggumononeofthemasses? ? Nowit’snownotsuch
asymmetric system andthe resultismorecomplicated. . Inthiscaseyoucanactually y solvedforthe
frequencies,sothatyoucantestthisperturbationmethodonsomethingthatyoucandoexplicitly. Let
m
2
=
m
1
=
m
be changedto
m
2
=
m
+
m
g
. Whatdoestheperturbation n method sayaboutthe
modiedfrequency?
M
0
=
m
0
0
m
;
M
1
=
0
0
0
m
g
;
K
1
=
0 0
0 0
(10
:
51)
Theequation(10.49)nowtellsyouthecorrectiontothefrequencyforeachmodeofoscillation.
x
0
=
1
1
!
!
2
1
=
(1 1)
k
1
m
0
0
0
m
g

1
1

(1 1)
m
0
0
m

1
1
k
1
m
m
g
2
m
(10
:
52)
x
0
=
1
1
!
!
2
1
=
(1  1)
k
1
+2
k
2
m
0
0
0
m
g

1
1

2
m
k
1
+2
k
2
m
m
g
2
m
(10
:
53)
Bothfrequenciesarelowered,andifyouaddsomemassthatiswhatyoushouldexpect.
Theexactsolutionisstatedinproblem10.25,allowingyoutochecktheresultsinthiscase.
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10|CoupledOscillators
351
Exercises
IntheexampleatEq.(10.1),takethecaseinwhich
k
1
=
k
3
=0andaddthetwoequations.What
doesthisresultmean?
WhathappensifyouaddthethreeequationsatFigure10.2?
JustbeforeEq.(10.5) thereisareferencetoEq.(3.1). Howdoesthisnewmaterialapply y tothat
equation?Where’sthematrix?
Verify the e result t of f Eq. (10.10) forthe e 22 2 case, , with h M M =
ab
cd
and N N having g components
e;f;g;h
.
ParallelingthecalculationinEq.(10.15),ndtheenergyforthecircuitofEq.(10.2)
AnswerthequestionintheparagraphimmediatelyafterEq.(10.40).
WhathappenstotheresultinEq.(10.24)ifallthenormalmodesolutions(thex
a
)aremultiplied
bytwo?
Explicitly verify Eq. (10.47) for the e 22 case. . That t is, take e the matrix =
ab
b c
with null
eigenvectorx=
e
f
andshowthatx
y
Hyiszeronomatterwhatyis. Note: : youdonothavetosolve
theeigenvalueequation;justuseit.
Translatetheequations(10.24)through(10.26)intothe11case.Thatis,whataretheseequations
forthesimpleharmonicoscillatorofchapterthree?
10 In the equation (10.28), look at the whole expression in n front t of the second
e
f
. What t is its
determinant?
11 Thereare twovectorswithcomponents(1
;
1
;
1), (0
;
1
;
1),and athirdis orthogonaltothe
rsttwo. Findit.
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10|CoupledOscillators
352
Problems
10.1 ExplicitlywriteoutthestepsofthederivationpresentedinEq.(10.10)forthe22case,with
MandNbeingarbitrary.etc.
10.2 Derivetheequations(10.11)and(10.12). Also,explicitlywriteouteachofthemterm-by-term
forthe22matrixcase.
10.3 ForthespecialcaseEq.(10.4),explicitlywriteouttheequations(10.16),(10.17),(10.18),and
(10.19)
10.4 If the mass matrix is proportional to
12
34
,and the scalarproductof two column matrices is
denedasinEq.(10.22),whichofthefourrequirementsforascalarproductareviolated? Whichof
thepropertiesofEq.(10.8)areviolated?
10.5 DothemanipulationsstartingatEq.(10.14)explicitlyforthe22case,verifyingtheresultsby
longhand.
10.6 Setuptheequationsforthefrequenciesofoscillationofthecoupledcircuits,Eq.(10.2),and
(a)ndthefrequenciesofoscillation.
(b)Inthespecialcasethatthemutualinductanceiszero,doyourresultsreducetothecorrectvalues?
(c)Ifthetwocapacitanceshavetheequalvaluesandtheself-inductancesdotoo,ndthefrequencies
andmodesofoscillation.
10.7 WriteoutthemanipulationsofEq.(10.48)explicitlyforthetwobytwocase,withH=
ab
cd
.
10.8 Ifyoudidn’tdoproblem7.16,nowisagoodtime.
10.9 VerifythatEq.(10.22)satisestherequirementsforascalarproduct.
10.10 Twopendulumshavethesamelengthandthesamemassattheirends. Theyare
attachedtoeachotherbyaverylightspringasinthedrawing,andthespringisrelaxed
whenboth massesarehangingvertically. . Setuptheequationsofmotionand d solvefor
thenormalmodesofoscillationandtheirfrequencies.
10.11 Usethegeneralexpression,Eq.(10.14),forenergyinacoupledharmonicsystem,
computetheenergyforthecoupledcircuitinEq.(10.2). Thisisnottheenergythatyou
normallyassociatewithcapacitorsandinductorsasdescribedinanybookonelectromag-
netism. There e you willseeonly
q
’s and
I
’s, with no
dI=dt
. Issomethingwrong? Gobackand d do
problem3.23toanswerthis.
10.12 Solveforthevibrationalmodesofoscillationofacarbondioxidemolecule.CO
2
islinear,making
themathabiteasierthanforawatermolecule. Forthisproblem,assumethatmotionisconnedto
onedimension. Ans:Onemodehasfrequency
!
2
=
k
2
m
o
+
m
c
 
m
o
m
c
)
10.13_RepeattheprecedingproblemfortheCO
2
molecule,butthistimethemoleculeisrotatingabout
anaxisthroughtheC-atomandperpendiculartotheaxisofthemolecule. Assumethattheangular
velocityisxedinspacewithmagnitude .
10.14RepeattheprecedingproblemfortheCO
2
molecule,withthemoleculerotatingaboutanaxis
throughtheC-atomandperpendiculartotheaxisofthemolecule. Nowaddatouchofrealismbynoting
thatangularmomentumisconserved:
L
z
=2
mr
2
!
z
,sothattheangularvelocityisnotconstant.
10|CoupledOscillators
353
10.15 Usethemethodsofsection10.4tosolvetheprobleminsection3.9leadingtoEq.(3.68).
10.16 Youcandotheexampleinequations(10.50)and(10.51)anotherway.Rewritethematricesin
amoresymmetricformbeforedoingtheperturbationcalculation.
M
0
=
m
+
1
2
m
g
0
0
m
+
1
2
m
g
;
M
1
=
1
2
m
g
0
0
m
g
;
K
1
=
0 0
0 0
Thesays thattheperturbationinvolvessubtractingsomemassfromoneandaddingittothe other.
Nowrepeattheperturbationanalysis. Showthattorstorderin
m
g
thetworesultsagree. Thisway
iseasier.
10.17 IntheexampleshowninFigure10.2,assumethatthemassesareallthesameasarethesprings.
Withoutsolvinganyequations,drawpicturesthatwilldescribethemodesofoscillation.Thisisinthe
plane,twodimensions,sotherearesixcoordinatesandsosixmodes. Inthreeofthecasesyoushould
beabletowritethefrequencycorrespondingtothemodeexactly. Intheotherthreecasesyouwould
havetodosomealgebratondtheresult,butyouneednot. Ifyoundsevenpossiblesolutions,you
shouldbeabletoshowthatoneisalinearcombinationoftwoothers.
10.18 Twopendulumsarecoupledwithalightspring. Thetwolengthsofthependulums
arethesame,butthemassesaredierent.Thespringisattachedatacommondistance
fromthesupportandthelengthsofthependulumsare
L
. Themassesare
m
1
and
m
2
,
andwhenthemassesarehangingverticallythespringisunstretched. Thesolutionsfor
thetwooscillationfrequenciesare
!
2
=
g=L
and
!
2
=(
g=L
)+(
k‘
2
=L
2
)where
is
thereducedmass. Therespectiveamplitudes s ofmotionforthetwomodesare(1 1 1)
and(1  
m
1
=m
2
). Analyzetheseproposedsolutionstoseeiftheyareplausible. Are
themodesorthogonal? [Thisisaneasydemonstration n tosetupand toseethatthemassmatrixis
importantinmakingthesemodesorthogonal.]
10.19_Derivethe solutionforthepreceding problem. . Set t up theequations ofmotion and nd the
frequenciesandmodesofoscillation.Ifyouthinkthatyou’regettinginvolvedinsolvingacomplicated
quarticequation,lookmoreclosely.
10.20 Threemasses,allequaltoeachother,areconnectedinastraightlinebetweenxed
wallsasinthedrawingFigure10.1. Allfourspringsareidenticaltoo. Withoutsolvingthe
equations,drawsketchesshowingwhatthenormalmodeswillbe.Asaguide,rememberthe
orthogonalityrelation,Eq.(10.22). Itwillhelpyouingettingtheshapesofthemodes.One
ofthemodesshouldbeveryeasytogureoutexactly.Anotheroneshouldbeeasytodoat
leastapproximately.Forthethird,remembertheorthogonality.
10.21 Astringof
N
equalmasses,
m
,areconnectedalongastraightlinebyequalsprings,
k
.Thelast
springsoneachendareattachedtowalls.(a)WritedowntheMandKmatricesforthissystem.(b)If
insteadofbeingattachedtowalls,thesystemis pulledaroundintoalargecirclesothatthemasses
arefreetomoveonlyalongthexedcircumferenceofthecircle. Thelastspringisthenattachedto
therstmass,makingasystemwithcircularsymmetry. WriteMandKnow. (c)Inbothcases,do
thesematricessatisfythereal-symmetricrequirementsofEq.(10.8)?
10.22_Insection3.11youndtheGreen’sfunctionforasimpleharmonicoscillator. Thereisasimilar
resultforcoupledharmonicoscillators,onlynowthefunctionwillbeamatrixoffunctions. Takethe
simplestspecialcaseoftwoequalmassesandthreesprings(symmetric)andkicktherstoneofthe
masses,asinEq.(3.72). Findtheresponseofthetwomassestothiskickandthenfollowthepattern
10|CoupledOscillators
354
ofthecalculationleadingtoEq.(3.73)togettheresultforanarbitraryforceappliedtotherstmass.
Youwillgetintegralsgivingyou
x
1
(
t
)and
x
2
(
t
).Repeatthisforaforceonthesecondmassandnally
writethewholeresultinmatrices
x(
t
)=
Z
t
1
dt
0
G(
t
t
0
)f(
t
0
)
whereGisa22matrix and fisatwoelementcolumn matrixrepresentingtheforces. . Doesyour
answermakesenseifthemiddlespringbecomeseitherveryweakorverystrong?
Ans:
G
11
=
1
2
m
1
!
sin
!t
+
1
!0
sin
!
0
t
10.23_Intheprecedingproblemyouusedthebasis
1
0
and
0
1
. Whatifyouswitchtothebasis
1
1
and
1
1
I.e.,maketherstkickequalonbothmasses(!
;
!). Makethesecondkickoppositeon
thetwomasses(!
;
). NowwhatisthematrixG?Again,doesyouranswermakesenseifthemiddle
springbecomeseitherveryweakorverystrong?
10.24 (a)Usetheresultofproblem10.22totodescribewhathappensifyouapplytheforce
F
0
sin
!t
totherstmass. (b)Whatistheresultiftheforceis
F
0
sin
!
0
t
? Here
!
and
!
0
arethetwonatural
frequenciesforthesystem.
10.25_Theexactsolutionforthefrequenciesfoundintheequations(10.52)and(10.53)are
!
2
=
h
(2
m
+
m
g
)(
k
1
+
k
2
)
q
m
2
g
(
k
1
+
k
2
)2+4
m
(
m
+
m
g
)
k
2
2
i
2
m
(
m
+
m
g
)
Verifythataseriesexpansionoftheexactsolutionreproducesthetwoapproximatesolutions.
10.26 WhatwillreplaceEq.(10.24)ifyouuse
e
i!t
insteadofcos
!t
andsin
!t
?
10.27_Likeproblem10.10,butwiththreependulumsinarow. Setupthegeneralequationsandthe
algebraicequationforthefrequencies. It’sacubicequationfor
!
2
,butyoushouldbeabletoseeone
solutioninstantly. Factorthatrootoutoftheequation. . Nowyoucanseetheshapeofanothermode
easily and thencompute itsfrequency. . Factorthat t rootoutofthe equation and youhavea linear
equationforthenal
!
2
.Whatistheshapeofthatmode,andarethethreemodesorthogonal?
10.28 Inproblem 10.19youmay havesolvedtheproblemexactly,butevenif youdidn’t,andifthe
massesarenotverydierent,useperturbationtheorytondtheoscillationfrequenciesapproximately.
10.29_Usethemethodsofsection10.4tosolveproblem10.22.
10.30 Drawpictures likethose in the gures 10.9{10.11 forthe cases oftwoand of three masses:
N
=2
;
3.
10.31 WhatisthetimedependenceinFigure10.11I.e.drawthepictureatlatertimes.
10.32 Solveproblem3.59. Arethemodesorthogonal? ? Isthescalarproductpositivedenite? ? Don’t
justassumethatitis;checkit.Ifyoudon’tknowanyfancytheoremstodothis,thenperhapsyoucan
computetheeigenvectorsandeigenvaluesofM,thendeducetheresultfromthere.
10.33 Todierentiateafunction numerically,usingonly the dataatdiscrete,equally spacedpoints,
thesimplestmethodestimatestheresultbythe\midpointrule":
f
0
(
x
+
1
2
h
)
f
(
x
+
h
f
(
x
)
=h
.
Youcanestimatethesecondderivativebyrstusingthismethodtoget
f
0
(
x
h=
2)and
f
0
(
x
+
h=
2).
Thengetthesecondderivativeat
x
by applyingthesamemidpointruleandusingthesetworesults.
10|CoupledOscillators
355
NowgobacktoEq.(10.30),divideitby
,andseewhatit(approximately)is. Youcaneventakethe
limitas
!0,andcomparethisresulttothewaveequationofchapterseven.
10.34_Generalizeproblem10.22tosolveanarbitrarycoupledoscillatingsystemwithanappliedforce:
Mx+Kx=f(
t
).Usethenotationofsection10.4,andtheanswerisclaimedtobe
x(
t
)=
Z
t
1
dt
0
G(
t
t
0
)f(
t
0
)
;
with
G(
t
)=
X
b
sin
!
b
t
!
b
x
y
b
Mx
b
x
b
x
y
b
Checkwhetherthisisplausible. Ifyou’renotsurewhatx
b
x
y
b
means(andwhyshouldyoube?),justput
itinfrontofthef(
t
0
)andseewhatitlookslikethen.
10.35Derivetheresultfortheprecedingproblem.Sortingoutthenotationisprobablythehardestpart.
Note:DonotassumethatthematrixMisdiagonal.Thisimpliesthatifyoustartfromrestandgivean
impulsetotherstmass,thenperhapsallthemasseswillimmediatelystartmoving. E.g.Eqs.(10.2).
Playwiththe22casersttoseewhatisgoingon,writingoutallthe
M
11
,
M
12
,
:::
explicitly. Go
backtothederivationstartingwith Eq.(3.71),only now
m
is thematrixMand1
=m
isthematrix
M
1
.
10.36 Usetheresultofproblems10.34and10.35toderivetheresultforproblem10.22.
10.37 (a) Ifyouhaven’tsolvedproblem3.80,doitnow. Evenifyouhave,lookatitagainandsee
howmucheasieritseems. Ihope. (b)Compareyourapproximateanswertotheexactanswerstated
inproblem10.25.
10.38 ExaminetheresultstatedinEq.(10.29)anddetermineifitisplausible.
10.39 Howdothe phasesofthemotions ofthe twomasses in Eq.(10.29)compare? Thequotient
x
1
(
t
)
=x
2
(
t
)cancelsboththetimedependenceandthedet’s. Plotthisrelativephaseversus .
10.40 Threemasses,all=
m
,arespacedequallyaroundawireandareabletoslide
alongthewire.Springsconnectthemasses,allthesame,andthemassescanoscillate.
Findthenormalmodesofthissystem. Didyoudoallofthepreliminaryexercisesat
theendofthischapter?
10.41 Returntoproblem3.60andanalyzeitusingthe tools ofthis chapter. . What
is theorthogonality relationship between thenormal modes,andwhatdo they look
like?
10.42 Insection10.6theboundaryconditionsonthemassesrequiredthatthereareanitenumberof
them,andtheendpointswereconnectedinsomeway. Nowassumethatthechainisliketheinnitely
longstringusedatthebeginningofchapter7. Itneverstops. . Thismeansthatyoucanmoreeasily
representthesolutionsasrunningwavessuchas
e
ikx
!t
. (a)Whatistherelationbetween
!
and
k
forthiscase? (b)GraphthephaseandgroupvelocitiesasinFigure10.12. (c)Whatconstraintsare
therenowonthepossiblevaluesof
!
and
k
?
NonlinearOscillations
.
Thesimpleharmonicoscillatorprovidesanexcellentmodelforoscillationsaboutequilibrium.A
massonaspring,apendulum,aplanetinanalmostcircularorbit,anatomvibratinginacrystal
:::
.
Evenwiththependulum however,itis apparentthatweareomittingsomething. . Inthatcasemore
thananyoftheothersIwasexplicitinapproximatingthesinebyitsargument.Intheseriesexpansion
sin
=
3
=
6+Idroppedeverythingbutthersttermandlookedbackonlyonce,insection
4.6. Isthisapproximationgood? Howgood?Insuchanapproximationaretherequalitativelydierent
phenomenathatIamignoringoraretheerrorsonlythoseofthenextdecimalplace?
Regardingthelastquestion,therearesomeofeach. Forthependulum,aslongastheoscillations
don’ttakeitnearthetop,thereareonlyquantitativedierences. Ifyouapplyanoscillatingforcetoit
howeverthereareother,radicallydierent,thingsthatcanhappen,startingwithhysteresisandonup
tochaos.
11.1ForcedPendulum,Qualitatively
Beforeattackingthismaterial,gobacktosection3.6onthe
forcedharmonicoscillator,andreaditagain.Especially,un-
derstandthegraphFigure3.5(a)andthematerialthatleads
uptoit.Whatfollowswillthenmakeawholelotmoresense.
Recallthephenomenonofresonancefromchapterthree,section 3.6. Youapplyanoscillating
forcetoasimple,undampedharmonicoscillator,andiftheforcingfrequency isclosetotheoscillator’s
naturalfrequency,yougetalargeresponse.Iftheforcingfrequencyisatthenaturalfrequencyyouget
aresponsethatrunsawayandgrowswithoutlimit.Whenyouincludesomefrictionittamestheresult
sothatitstaysnite,asinthisgraphandthegraphfromFigure3.5(a).Here,
b=m!
0
=0
:
1
!
0
Fig.11.1
Ifthisoscillatorisapendulum,thenthisanalysisisgoodonlytotheextentthatthesmallangle
approximationisgood,allowingthependulumtobeapproximatedbyasimpleharmonicoscillator. If
thisgraphrepresentsthemaximumangleofoscillationforthependulumandiftheverticalscalemakes
thepeakofthegraph30
thenthisholdsnosurprisesbecause30
=
=
6=0
:
5236,andthatisstill
veryclosetosin30
=0
:
5. Whatifthereislessdamping,sothatthepeakofthisresonancecurve
wouldtakeyouto90
or150
? Thenthissmallangleapproximationbreaksdown. Moreimportantly,
itbreaksdownnotonlyquantitatively,butqualitatively.Newphenomenaappear,onesthatdon’tshow
upinthesmallangleapproximation.
Thenaturalperiodofapendulumchangesasthemaximumangleincreases.Whenthemaximum
anglegets upto179
,itstopsandwaits whileitconsiderswhethertoreturnortogooverthetop.
Theperiodisthenmuchlongeratthislargeangleofoscillation.ThisisfullysolvedinEq.(4.63).
Startthisanalysisbypushingthependulumatafrequencywellabovethenaturalfrequency,so
thatyou’retotherightofthepeakin the curve ofFigure 11.1 Theresponseisthenproportionate,
356
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